A calculus problem by Guilherme Dela Corte

$\begin{aligned} I & = \int_1^{\sqrt e} x \ln x \ dx & \small \color{#3D99F6} \text{By integration by parts} \\ & = \frac {x^2\ln x}2 \ \bigg|_1^{\sqrt e} - \int_1^{\sqrt e} \frac {x^2}{2x} \ dx \\ & = \frac {x^2\ln x}2 - \frac {x^2}4 \ \bigg|_1^{\sqrt e} \\ & = \frac 14 \end{aligned}$

$\implies I^{-1} = \boxed{4}$

We know that $\left (\frac{x^2}{2}\ln x \right )' = x \ln x + \frac{x^2}{2} \cdot \frac{1}{x}$ . Integrating both sides to this equation, it yields:

$\frac{x^2}{2}\ln x + C_0= \int \left (x \ln x \right ) \: \partial x + \int \frac{x}{2} \: \partial x \Leftrightarrow \boxed{\displaystyle \int \left (x \ln x \right ) \: \partial x = \frac{x^2}{2}\ln x - \frac{x^2}{4} + C}$

Thus, $\displaystyle \int_{1}^{\sqrt{e}} \left (x \ln x \right ) \: \partial x = \left [\frac{x^2}{2}\ln x - \frac{x^2}{4} + C \right ]_{1}^{\sqrt{e}}=\left (\frac{e}{4} - \frac{e}{4} \right ) - \left ( 0 - \frac{1}{4} \right ) = \frac{1}{4}$ , and we have finally that

$\left ( \displaystyle \int_{1}^{\sqrt{e}} \left (x \ln x \right ) \: \partial x \right )^{-1} = \; \; {\color{#D61F06} \boxed{4.}}$